Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long int pow_mod(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
struct seg_node {
  int cnt;
  seg_node *lft, *rgt;
  seg_node(int cnt, seg_node *lft, seg_node *rgt)
      : cnt(cnt), lft(lft), rgt(rgt) {}
  seg_node *insert(int l, int r, int val);
};
seg_node *null = new seg_node(0, NULL, NULL);
seg_node *seg_node::insert(int l, int r, int idx) {
  if (l == r) {
    return new seg_node(this->cnt + 1, null, null);
  }
  int mid = (l + r) >> 1;
  if (idx <= mid) {
    return new seg_node(this->cnt + 1, lft->insert(l, mid, idx), this->rgt);
  } else {
    return new seg_node(this->cnt + 1, this->lft, rgt->insert(mid + 1, r, idx));
  }
}
seg_node *tree[100005];
bool visited[100005];
vector<int> adj[100005];
int idx[100005];
vector<int> pop_sz[100005];
int tin[100005], tout[100005];
int tme = 0;
int dp[25][100005];
int len;
void __dfs(int nd, int par) {
  tin[nd] = tme++;
  dp[0][nd] = par;
  for (int i = 1; i < len; ++i) {
    dp[i][nd] = dp[i - 1][dp[i - 1][nd]];
  }
  for (int i = 0; i < (int)(adj[nd].size()); ++i) {
    int to = adj[nd][i];
    if (to != par) {
      __dfs(to, nd);
    }
  }
  tout[nd] = tme++;
}
bool isParent(int par, int ch) {
  return (tin[par] <= tin[ch] && tout[ch] <= tout[par]);
}
int lca(int a, int b) {
  if (isParent(a, b)) {
    return a;
  }
  if (isParent(b, a)) {
    return b;
  }
  for (int i = len - 1; i >= 0; --i) {
    if (!isParent(dp[i][a], b)) {
      a = dp[i][a];
    }
  }
  return dp[0][a];
}
void init_lca(int n) {
  len = 1;
  while ((1 << len) <= n) {
    ++len;
  }
  __dfs(1, 1);
}
void build_tree(int nd, int par, int n) {
  tree[nd] = tree[par];
  visited[nd] = true;
  for (int i = 0; i < (int)(pop_sz[nd].size()); ++i) {
    int to = pop_sz[nd][i];
    tree[nd] = tree[nd]->insert(1, n, to);
  }
  for (int i = 0; i < (int)(adj[nd].size()); ++i) {
    int to = adj[nd][i];
    if (!visited[to]) {
      build_tree(to, nd, n);
    }
  }
}
void solve(int L, int R, seg_node *nd1, seg_node *nd2, seg_node *nd3,
           seg_node *nd4) {
  int tmp = nd1->cnt + nd2->cnt;
  tmp = tmp - nd3->cnt - nd4->cnt;
  if (tmp) {
    if (L == R) {
      printf("%d", L);
      printf(" ");
      return;
    }
    int mid = (L + R) >> 1;
    solve(L, mid, nd1->lft, nd2->lft, nd3->lft, nd4->lft);
    solve(mid + 1, R, nd1->rgt, nd2->rgt, nd3->rgt, nd4->rgt);
  }
}
int main() {
  int n, m, q, u, v, a;
  scanf("%d", &n);
  scanf("%d", &m);
  scanf("%d", &q);
  for (int i = 0; i < n - 1; ++i) {
    scanf("%d", &u);
    scanf("%d", &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= m; ++i) {
    scanf("%d", &u);
    pop_sz[u].push_back(i);
  }
  tree[0] = null;
  tree[0]->lft = null;
  tree[0]->rgt = null;
  memset(visited, false, sizeof(visited));
  build_tree(1, 0, m);
  init_lca(n);
  while (q--) {
    scanf("%d", &u);
    scanf("%d", &v);
    scanf("%d", &a);
    seg_node *nd1 = tree[u];
    seg_node *nd2 = tree[v];
    int _lca = lca(u, v);
    seg_node *nd3 = tree[_lca];
    seg_node *nd4 = tree[dp[0][_lca]];
    if (_lca == 1) {
      nd4 = tree[0];
    }
    int tmp = nd1->cnt + nd2->cnt - nd3->cnt - nd4->cnt;
    a = min(a, tmp);
    int L = 1, R = m;
    printf("%d", a);
    printf(" ");
    while (L < R) {
      tmp = nd1->lft->cnt + nd2->lft->cnt;
      tmp = tmp - nd3->lft->cnt - nd4->lft->cnt;
      int mid = (L + R) >> 1;
      if (tmp >= a) {
        nd1 = nd1->lft;
        nd2 = nd2->lft;
        nd3 = nd3->lft;
        nd4 = nd4->lft;
        R = mid;
      } else {
        solve(L, mid, nd1->lft, nd2->lft, nd3->lft, nd4->lft);
        nd1 = nd1->rgt;
        nd2 = nd2->rgt;
        nd3 = nd3->rgt;
        nd4 = nd4->rgt;
        L = mid + 1;
        a -= tmp;
      }
    }
    if (L) {
      solve(L, L, nd1, nd2, nd3, nd4);
    }
    printf("\n");
  }
  return 0;
}